On the absolute continuity and the Riemann integrability of a particular functions.

We consider $f\in L^1[a,b]$, where $F$ is a finite interval of $\mathbb{R}$. The function $F\colon [a,b]\to\mathbb{R}$ defined as $$F(x):=\int_{[a,x]}f\;d\lambda\quad(x\in [a,b])$$it is called integral function with respect to the function $f$. It is easily shown that $F$ is absolutely continuous on $[a,b]$.

Now, we extend the definition of $F$ as follows:

$$F(x):=F(b)\quad\text{for}\quad x>b$$

Now, for $n\in \mathbb{N}$ we consider the function $$G_n(x):=\frac{F_1\left(x+\frac{1}{n}\right)-F(x)}{\frac{1}{n}}\quad x\in [a,b]$$

It seems completely obvious to me that $G_n$ is also absolutely continuous on $[a,b]$ for all $n\in\mathbb{N}$, but I can't show it.

My attempt

The first step consists in proving that the extension of $F$, i.e. the function $F_1\colon [a,\infty)\to\mathbb{R}$ defined as $$F_1=F\;\text{on}\; [a,b]\quad F_1=F(b)\;\text{on}\; (b,\infty)$$ is absolutely continuous on $[a,\infty)$. I consider, for semplicity, one closed interval $[a_1, b_1]\subseteq [a,\infty)$. Let $\varepsilon>0$ fixed and we suppose that $(b_1-a_1)<\delta$. If $[a_1, b_1]\subseteq [a,b]$, then $$\lvert F_1(b_1)- F_1(a_1)\rvert =\lvert F(b_1)-F(a_1)\rvert < \varepsilon,$$ this holds for the absolute continuity of $F$ on $[a,b]$. If $a_1<b, b_1> b$ we have that $$\lvert F_1(b_1)-F_1(a_1)\rvert=\lvert F(b)-F(a_1)\lvert<\varepsilon$$ in fact $(b-a_1)<(b_1-a_1)<\delta$ and also this holds for the aboslute continuity of $F$ on $[a,b]$. Finally, if $a_1> b, b_1> b$ we have that $$\lvert F_1(b_1)-F_1(a_1)\rvert=\lvert F(b)-F(b)\rvert=0<\varepsilon$$This proves that $F_1$ is absolutely continuous on $[a,\infty)$. Now we consider for $n\in\mathbb{N}$ the function $$x\mapsto F_1\left(x+\frac{1}{n}\right)\quad (x\in [a,b])$$and we prove absolute continuity on $[a,\infty)$. We fix $\varepsilon > 0$ and we consider the closed interval $[a_1,b_1]\subseteq [a,\infty)$ such that $(b_1-a_1)<\delta$. Now, we evaluete the quantity $$\left\lvert F_1\left(b_1+\frac{1}{n}\right)-F_1\left(a_1+\frac{1}{n}\right)\right\rvert \tag1$$we note that $(1)$ is less than $\varepsilon$ as $(b_1+\frac{1}{n})-(a_1+\frac{1}{n})=(b_1-a_1)<\delta$ and $F_1$ is absolutely continuous on $[a,\infty).$

Then the function $x\mapsto F_1(x+\frac{1}{n})$ is absolutely continuous on $[a,\infty)$ and so, in particular, it is in $[a,b]$. Then the function $G_n$ are in $AC[a,b]$ as a combination of $AC[a,b]-$functions.

Is this reasoning sufficient?